Question: Let $h$ be a vector-valued function defined by $h(t)=(-t^5-6,4t^4+2t+1)$. Find $h$ 's second derivative $h''(t)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\left(-20t^3,48t^2\right)$ (Choice B) B $\left(t^3,4t^2\right)$ (Choice C) C $-5t^4+16t^3+2$ (Choice D) D $\left(-5t^4,16t^3+2\right)$
Solution: We are asked to find the second derivative of $h$. This means we need to differentiate $h$ twice. In other words, we differentiate $h$ once to find $h'$, and then differentiate $h'$ (which is a vector-valued function as well) to find $h''$. Recall that $h(t)=(-t^5-6,4t^4+2t+1)$. Therefore, $h'(t)=\left(-5t^4,16t^3+2\right)$. Now let's differentiate $h'(t)=\left(-5t^4,16t^3+2\right)$ to find $h''$. $h''(t)=\left(-20t^3,48t^2\right)$ In conclusion, $h''(t)=\left(-20t^3,48t^2\right)$.